winnard2008 Posted September 22, 2008 Share Posted September 22, 2008 Hi Guys, For some reason my login script does not check the values in the database for a match and simply logs you in regardless of what you type in the username and password fields. Any help would be great. Here is the code <?PHP // Send nothing to the Web Browser until the Session_start() Line. // Check if the form has been submitted. if (isset($_POST['submitted'])) { require_once ('xxxxxxxxxx'); // Connect to the database. $errors = array(); // Initialise Error Array. // Check for a UserName. if(empty($_POST['UserName'])) { $errors[] = '<p>You forgot to enter a User Name.</p>'; } else { $u = ($_POST['UserName']); } //Check for a Password. if (empty($_POST['Password'])) { $errors[] = '<p>You forgot to enter a Password</p>'; } else { $p = ($_POST['Password']); } if (empty($errors)) { // If everything's OK. // Retrieve the UserID and UserName for login combination. $query = "SELECT UserID FROM Users WHERE UserName='$u' AND Password=SHA('$p')"; $result = @mysql_query ($query); $row = mysql_fetch_array ($result, MYSQL_NUM); if ($row)) { // A record was pulled from the database. //Start the session for this user. session_start(); $_SESSION['UserID'] = $row[0]; $_SESSION['UserName'] = $row[1]; } //Redirect the user to the admin home panel. if (!headers_sent()) { header ('Location: index.html'); exit(); // Quit the script. } else { // No record matched the query. $errors[] = 'The User Name and Password entered do not match our records.'; $errors[] = mysql_error() . '<br /><br />Query: ' . $query; } } mysql_close(); // Close the database connection. } else { // Form has not been submitted. $errors = NULL; } // End of the main submit conditional. // Begin the page. ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title></title> <link href="css/styles2.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="wrapper"> <div id="header"> </div> <div id="right_column"> <?PHP if (!empty($errors)) { // Print any error messages. echo '<h1>Error!</h1> <p>The following error(s) occurred:<br />'; foreach ($errors as $msg) { echo " - $msg<br />\n"; } echo '</p><p>Please try again.</p>'; } // Create the form. ?><br /><br /> <div align="center"><h2>Login</h2></div> <form id="form" action="login.php" method="post"> <p>Username: <input type="text" name="UserName" size="10" maxlength="10" class="textarea" /></p><br /> <p>Password: <input type="password" name="Password" size="10" maxlength="10" class="password"/></p><br /> <p><input type="submit" name="submit" value="Login" class="button"/></p><br /> <input type="hidden" name="submitted" value="TRUE" /> </form> </div> Quote Link to comment Share on other sites More sharing options...
Maq Posted September 22, 2008 Share Posted September 22, 2008 Have you tried echoing out your values to make sure they're correct? Quote Link to comment Share on other sites More sharing options...
Brian W Posted September 22, 2008 Share Posted September 22, 2008 $row = mysql_fetch_array ($result, MYSQL_NUM); if ($row)) { // A record was pulled from the database. //Start the session for this user. session_start(); $_SESSION['UserID'] = $row[0]; $_SESSION['UserName'] = $row[1]; } I think your problem is within these lines... but I cant put my finger on it. Trouble shoot please./ disable your redirect for a moment and echo $u, $p, $row[0], and $row[1] if they are all working properly, than I'll keep looking into your code, other wise maybe we can rework the syntax to get what you want. Quote Link to comment Share on other sites More sharing options...
winnard2008 Posted September 22, 2008 Author Share Posted September 22, 2008 Hi I am getting an error to do with the query. When I echo the stuff you wanted me to echo it was bringing up a query error and not displaying the things I echo'd out. Quote Link to comment Share on other sites More sharing options...
Maq Posted September 22, 2008 Share Posted September 22, 2008 Try: $query = "SELECT UserID FROM Users WHERE UserName='" . $u . "' AND Password=SHA('" . $p . "')"; Quote Link to comment Share on other sites More sharing options...
Brian W Posted September 22, 2008 Share Posted September 22, 2008 Does the query error tell you anything? like does it have a excerpt from the syntax or the SQL? Can't think of whether it will or not, but worth asking. Any way, try $query = "SELECT UserID FROM Users WHERE UserName='".$u."' AND Password='".SHA($p)."'"; LOL, or what the other guy beat me too. I didn't know you could SHA with SQL... let me know who's works. (if either) That might actually do the trick... but i'm not that great at the whole sql thing yet. Quote Link to comment Share on other sites More sharing options...
Maq Posted September 22, 2008 Share Posted September 22, 2008 SHA($p) What is SHA? A function or hard coded text? Quote Link to comment Share on other sites More sharing options...
Brian W Posted September 22, 2008 Share Posted September 22, 2008 SHA is like md5() or whatever, it hashes the string like encrypting. SHA was developed by some government branch related to national security. Quote Link to comment Share on other sites More sharing options...
Maq Posted September 22, 2008 Share Posted September 22, 2008 Well is it part of PHP? Quote Link to comment Share on other sites More sharing options...
winnard2008 Posted September 22, 2008 Author Share Posted September 22, 2008 Hi Guys, Thanks for all your efforts but whilst I have changed the query and I am getting no syntax errors, it still is logging me in under any username passwords. I must admit I am very new to PHP and knowing where to echo out the values and error reporting isnt my strongest point. Maybe you could guide in this area and I will be able to find out if there are errors anywhere else. Cheers Danny Quote Link to comment Share on other sites More sharing options...
Maq Posted September 22, 2008 Share Posted September 22, 2008 Sure, for error reporting put this in. It will show errors only for the session. // Send nothing to the Web Browser until the Session_start() Line. // Check if the form has been submitted. error_reporting(E_ALL); ini_set('error_reporting', E_ALL); if (isset($_POST['submitted'])) { Let us know what errors you get. Quote Link to comment Share on other sites More sharing options...
winnard2008 Posted September 22, 2008 Author Share Posted September 22, 2008 I don't get a single error????????????? I think I am just going to re-write this code because I cannot understand what is wrong with it. When I log in, it directs to the index page. Even if the username and password are completely bogus. I really dont understand it. Quote Link to comment Share on other sites More sharing options...
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